248	Week 5: Torque and Rotation in One Dimension

F

pivot

F

F

Figure 67: What direction does one expect the spool in each of the figures to roll (or will it roll at all)?

5.5: Torque and the Center of Gravity

We will often wish to solve problems involving (for example) rods pivoted at one end swinging down under the influence of near-Earth gravity, or a need to understand the trajectory and motion of a spinning basketball. To do this, we need the idea of the center of gravity of a solid object. Fortunately, this idea is very simple:

The center of gravity of a solid object in an (approximately) uniform near-Earth gravitational field is located at the center of mass of the object. For the purpose of evaluating the torque and angular motion or force and coordinate motion of center of mass, we can consider that the entire force of gravity acting on the object is equivalent to the the force that would be exerted by the entire mass located as a point mass at the center of gravity.

The proof for this is very simple. We’ve already done the Newtonian part of it – we know that the total force of gravity acting on an object makes the center of mass move like a particle with the same mass located there. For torque, we recall that:

τ = rF = r F

(492)

If we consider the torque acting on a small chunk of mass in near-Earth gravity, the force (down) acting on that chunk is:

dFy = −gdm	(493)
The torque (relative to the pivot) is just:
dτ = −gr dm	(494)
or
τ = −g Z0L r dm = −M grcm	(495)

where rcm is (the component of) the position of the center of mass of the object perpendicular to gravity. The torque due to gravity acting on the object around the selected pivot axis is the same

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as the torque that would be produced by the entire weight of the object pulling down at the center of mass/gravity. Q.E.D.

Note that this proof is valid for any shape or distribution of mass in a uniform gravitational field, but in a non-uniform field it would fail and the center of gravity would be di erent from the center of mass as far as computing torque is concerned. This is the realm of tides and will be discussed more in the week where we cover gravity.

Example 5.5.1: The Angular Acceleration of a Hanging Rod

L

Figure 68: A rod of mass M and length L is suspended/pivoted from one end. It is pulled out to some initial angle θ0 and released.

This is your first example of what we will later learn to call a physical pendulum. A rod is suspended from a pivot at one end in near-Earth gravity. We wish to find the angular acceleration α as a function of θ. This is basically the equation of motion for this rotational system – later we will learn how to (approximately) solve it.

As shown above, for the purpose of evaluating the torque, the force due to gravity can be considered to be M g straight down, acting at the center of mass/gravity of the rod (at L/2 in the middle). The torque exerted at this arbitrary angle θ (positive as drawn, note that it is swung out in the counterclockwise/right-handed direction from the dashed line) is therefore:

τ = rFt = −	M gL	sin(θ)	(496)
	2

It is negative because it acts to make θ smaller ; it exerts a “twist” that is clockwise when θ is counterclockwise and vice versa.

From above, we know that I = M L2/3 for a rod pivoted about one end, therefore:

τ = −M2gL sin(θ)

d2θ α = dt2

independent of the mass!

=		M L2			α = Iα	or:
			3
			3g
=	−			sin(θ)		(497)
			2L

5.6: Solving Newton’s Second Law Problems Involving Rolling

One of the most common applications of one dimensional torque and angular momentum is solving rolling problems. Rolling problems include things like:

250	Week 5: Torque and Rotation in One Dimension

•A disk rolling down an inclined plane.

•An Atwood’s Machine, but with a massive pulley.

•An unwinding spool of line, either falling or being pulled.

These problems are all solved by using a combination of Newton’s Second Law for the motion of the center of mass of the rolling object (if appropriate) or other masses involved (in e.g. Atwood’s Machine) and Newton’s Second Law for 1 dimensional rotation, τ = Iα. In general, they will also involve using the rolling constraint:

If a round object of radius r is rolling without slipping, the distance x it travels relative to the surface it is rolling on equals rθ, where θ is the angle it rolls through.

That is, all three are equivalently the “rolling constraint” for a ball of radius r rolling on a level floor, started from a position at x = 0 where also θ = 0:

x	=	rθ	(498)
v	=	rω	(499)
a	=	rα	(500)
			(501)

These are all quite familiar results – they look a lot like our angular coordinate relations – but they are not the same thing! These are constraints, not coordinate relations – for a ball skidding along the same floor they will be false, and for certain rolling pulley problems on your homework you’ll have to figure out one appropriate for the particular radius of contact of spool-shaped or yo-yo shaped rolling objects (that may not be the radius of the object!)

It is easier to demonstrate how to proceed for specific examples than it is to expound on the theory any further. So let’s do the simplest one.

Example 5.6.1: A Disk Rolling Down an Incline

r

	N
fs	m
	φ
	mg

φ

Figure 69: A disk of mass M and radius r sits on a plane inclined at an angle φ with respect to the horizontal. It rolls without slipping down the incline.

In figure 69 above, a disk of mass M and radius r sits on an inclined plane (at an angle φ) as shown. It rolls without slipping down the incline. We would like to find its acceleration ax down the incline, because if we know that we know pretty much everything about the disk at all future times that it remains on the incline. We’d also like to know what fs (the force exerted by static friction)

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when it is so accelerating, so we can check to see if our assumption of rolling without slipping is justified. If φ is too large, we are pretty sure intuitively that the disk will slip instead of roll, since if φ ≥ π/2 we know the disk will just fall and not roll at all.

~

As always, since we expect the disk to physically translate down the incline (so ~a and F tot will point that way) we choose a coordinate system with (say) the x-axis directed down the inline and y directed perpendicular to the incline.

Since the disk rolls without slipping we know two very important things:

1)The force fs exerted by static friction must be less than (or marginally equal to) µsN . If, in the end, it isn’t, then our solution is invalid.

2)If it does roll, then the distance x it travels down the incline is related to the angle θ it rolls through by x = rθ. This also means that vx = rω and ax = rα.

We now proceed to write Newton’s Laws three times: Once for the y-direction, once for the x-direction and once for one dimensional rotation (the rolling). We start with the:

Fy = N − mg cos(φ) = may = 0

(502)

which leads us to the familar N = mg cos(φ).

Next:

Fx = mg sin(φ) − fs	=	max	(503)
τ = rfs	=	Iα	(504)

Pay attention here, because we’ll do the following sort of things fairly often in problems. We use

I = Idisk = 1 mr2	and α = ax/r and divide the last equation by r on both sides. This gives us:
2
	mg sin(φ) − fs	=	max			(505)
	fs	=	1		max	(506)
				2

If we add these two equations, the unknown fs cancels out and we get:

	mg sin(φ) =			3	max	(507)
				2
	or:	2
	ax =		g sin(φ)			(508)
		3

We can then substitute this back into the equation for fs above to get:

	fs =		1	max =			1	mg sin(φ)	(509)
			2				3
In order to roll without slipping, we know that fs ≤ µsN or
1		mg sin(φ) ≤ µsmg cos(φ)							(510)
	3
or			1
			µs ≥			tan(φ)			(511)
					3

If µs is smaller than this (for any given incline angle φ) then the disk will slip as it rolls down the incline, which is a more di cult problem.

We’ll solve this problem again shortly to find out how fast it is going at the bottom of an incline of length L using energy, but in order to this we have to address rotational energy. First, however, we need to do a couple more examples.

252 Week 5: Torque and Rotation in One Dimension

M
T	r T
1	2

T1

T2

m1

m2

m 1 g

m 2 g

Figure 70: Atwood’s Machine, but this time with a massive pulley of mass M and radius R. The massless unstretchable string connecting the two masses rolls without slipping on the pulley, exerting a torque on the pulley as the masses accelerate to match. Assume that the pulley has a moment of inertia βM r2 for some β. Writing it this way let’s us use β ≈ 12 to approximate the pulley with a disk, or use observations of a to measure β and hence tell something about the distribution of mass in the pulley!

Example 5.6.2: Atwood’s Machine with a Massive Pulley

Our solution strategy is almost identical to that of our first solution back in week 1 – choose an ”around the corner” coordinate system where if we make moving m2 down “positive”, then moving m1 up is also “positive”. To this we add that a positive rotation of the pulley is clockwise, and that the rolling constraint is therefore a = rα.

Now we again write Newton’s Second Law once for each mass and once for the rotating pulley (as τ = Iα):

m2g − T2	=	m2a			(512)
T1 − m1g	=	m1a			(513)
τ = rT2 − rT1	=	βM r2	a	= Iα	(514)
			r

Divide the last equation by r on both sides, then add all three equations to eliminate both

	unknown tensions T1 and T2. You should get:
	(m2 − m1)g = (m1 + m2 + βM )a			(515)
	or:	(m2 − m1)g
	a =			(516)
		(m1 + m2 + βM )

This is almost like the previous solution – and indeed, in the limit M → 0 is the previous solution – but the net force between the two masses now must also partially accelerate the mass of the pulley. Partially because only the mass near the rim of the pulley is accelerated at the full rate a – most of the mass near the middle of the pulley has a much lower acceleration.

Note also that if M = 0, T1 = T2! This justifies – very much after the fact – our assertion early on that for a massless pulley, the tension in the string is everywhere constant. Here we see why that is true – because in order for the tension in the string between two points to be di erent, there has to be some mass in between those points for the force di erence to act upon! In this problem, that mass is the pulley, and to keep the pulley accelerating up with the string, the string has to exert a torque on the pulley due to the unequal forces.